Log_2(x-2)+log_2(x-3)=1

To solve the given logarithmic equation \( \log_2(x-2) + \log_2(x-3) = 1 \), we can use the properties of logarithms.

1. Combine the two logarithms using the product rule of logarithms: \[ \log_2[(x-2)(x-3)] = 1 \]

2. Remove the logarithm by converting it to exponential form: \[ 2^1 = (x-2)(x-3) \]

3. Simplify the exponential equation: \[ 2 = x^2 - 5x + 6 \] \[ x^2 - 5x - 4 = 0 \]

4. Factor the quadratic equation: \[ (x - 4)(x + 1) = 0 \]

5. Solve for \( x \): \[ x = 4 \] or \( x = -1 \)

Therefore, the solutions to the equation \( \log_2(x-2) + \log_2(x-3) = 1 \) are \( x = 4 \) or \( x = -1 \).